Question

If the standard molar enthalpies of formation of \( C_2H_5OH(l)) \), \( CO_2(g) \), and \( H_2O(l) \) are -278, -394, and -286 kJ mol\(^{-1}\) respectively, the standard enthalpy of combustion of the reaction \[ C_2H_5OH(l) + 3O_2(g) \rightarrow 2CO_2(g) + 3H_2O(l) \] in kJ mol\(^{-1}\) is:

• A. \(-1668\) kJ mol\(^{-1}\)
• B. \(-756\) kJ mol\(^{-1}\)
• C. \(-1258\) kJ mol\(^{-1}\)
• D. \(-1368\) kJ mol\(^{-1}\)

Hint:

The standard enthalpy of combustion is calculated by subtracting the sum of the enthalpies of formation of reactants from the sum of the enthalpies of formation of the products.

Answer 1

The Correct Option is D

The standard enthalpy of combustion for the given reaction is given by the equation: \[ \Delta H_{\text{comb}}^\circ = \sum \Delta H_f^\circ (\text{products}) - \sum \Delta H_f^\circ (\text{reactants}) \] Substituting the known values for enthalpies of formation: \[ \Delta H_{\text{comb}}^\circ = [2 \times (-394) + 3 \times (-286)] - [(-278) + 3 \times (0)] \] \[ \Delta H_{\text{comb}}^\circ = [-788 - 858] - [-278] \] \[ \Delta H_{\text{comb}}^\circ = -1646 + 278 \] \[ \Delta H_{\text{comb}}^\circ = -1368 \, \text{kJ/mol}. \] Thus, the standard enthalpy of combustion of the reaction is -1368 kJ/mol.